<u>Answer:</u> The expression for 
is written below.
<u>Explanation:</u>
Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as
For a general chemical reaction:
The expression for is written as:
The partial pressure for solids and liquids are taken as 1.
For the given chemical equation:
The expression for for the following equation is:
The partial pressure of 
will be 1 because it is solid.
So, the expression for now becomes:
Hence, the expression for is written above.
Answer:
Explanation:
Hello!
In this case, since the chemical reaction between copper and nitric acid is:
By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:
However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:
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1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba
<span>This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. </span>
<span>0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4</span>
We calculate for the number of moles of water given its mass by dividing the given mass by the molar mass.
n water = (36.04 g) / (18 g/mol)
n water = 2 mols
From the given balanced equation, every 6 moles of water produced will require 7 moles of oxygen.
n oxygen = (2 mols H2O) x (7 moles O2 / 6 moles H2O)
n oxygen = 2.33 mols O2
Answer:
Explanation:
Hello!
In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:
Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:
Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:
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